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Questions about the branch of algebra that deals with groups.
5
votes
2
answers
425
views
Centre of group with deficiency at least two (Progress on Murasugi's conjecture)
In 1965, Murasugi [1] conjectured that any finitely presented group with deficiency at least two has trivial centre. The year before, he had proved it true for one-relator groups, and in [1] he proved …
8
votes
1
answer
348
views
Finite two-relator groups and quotients of knot groups
Let $G$ be a one-relator group $\langle A \mid R = 1 \rangle$. Then clearly $G$ is finite if and only if it is cyclic of finite order, i.e. can be given by a presentation $\langle a \mid a^n = 1 \rang …
5
votes
Prove these are not surface groups
Here's an argument that uses only basic combinatorial group theory (Reidemeister-Schreier).
Let $n \geq 2$, and let $G = \langle a_1, b_1, \dots, a_g, b_g | [a_1, b_1]^n [a_2, b_2] \cdots [a_g, b_g] \ …
9
votes
3
answers
490
views
Residually solvable Bianchi groups
Let $d$ be a square-free positive integer, and let $\mathcal{O}_d$ be the ring of integers of the quadratic imaginary number field $\mathbb{Q}(\sqrt{-d})$. Consider the Bianchi group $\Gamma_d = \oper …
7
votes
Presentation of special linear group over localizations of the integers
In my recent paper (arXiv:2401.08146), I give a new presentation for $\operatorname{SL}_2(\mathbf{Z}[\frac{1}{2}])$. This group is generated by the two matrices
$$
A = \begin{pmatrix}1 & 0 \\ 1 & 1\en …
20
votes
1
answer
977
views
Proof of CFSG assuming every simple group is two-generated
It is well-known that one of the corollaries of the classification of finite simple groups (CFSG) is that every finite simple group can be generated by two elements. In a comment on an answer to an ol …
24
votes
Recognizing free groups
As indicated in the comments, it's undecidable in general to take as input a finite presentation of a group and try to output whether or not this group is free or not. This is a direct consequence of …
10
votes
Analogous results in geometric group theory and Riemannian geometry?
Here is a very classical example. As stated in the comments, Gromov was an early proponent of importing ideas from geometry to group theory, but already thirty years earlier there was work in this dir …
18
votes
1
answer
750
views
Is solvability semi-decidable?
Let $G = \langle A \mid R \rangle$ be a finitely presented group, given by a finite presentation. If $G$ is abelian, then we can verify this fact: simply verify the fact that $[a, b] = 1$ for all gene …
11
votes
Accepted
Cohomology of $\operatorname{GL}_3(\mathbb{F}_2)$
As you mention in your update, you have a general answer, but if you want a concrete answer for the low-dimensional integral cohomology of $G = \operatorname{GL}(3,2)$ (or any other finite group!), yo …
9
votes
Accepted
Is the isomorphism problem solvable for torsion-free groups?
Novikov's centrally-symmetric group $\mathfrak{A}_P$ is a torsion-free group with undecidable word problem, constructed in [1]. Novikov did not prove it is torsion-free but, as Adian points out in [Ad …
8
votes
2
answers
480
views
Subgroup membership problem in simple groups
Let $G$ be a finitely presented simple group. By Kuznetsov (1958), $G$ has decidable word problem. However, by Scott [1], $G$ may have undecidable conjugacy problem. Is anything known about other deci …
7
votes
Subgroup membership problem in simple groups
After some digging, I was able to find that the answer to my question exists: the problem can be undecidable. Rattaggi, in an unpublished manuscript (available here), proved that there exists a finite …
3
votes
Which groups are LERF?
Polycyclic groups are LERF, by Mal'cev 1948. In particular, all nilpotent and all abelian groups are LERF.
As mentioned in the comments, as not all one-relator groups are residually finite, not all on …
6
votes
Accepted
Is there a residually nilpotent one-relator group that is not residually a finite p-group fo...
The answer is yes, such a group exists: this is the main result of [1]. In fact, one can take the Baumslag-Solitar group $\operatorname{BS}(p^r, -p^r) = \langle a, b \mid ba^{p^r}b^{-1} = a^{-p^r} \ra …